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For all $n$, let $\gamma_k^n$ bet the tautological bundle over $Gr_k(\mathbb R^n)$, i.e. $$\gamma_k^n=\{(V,v):V\in Gr_k(\mathbb R^n), v\in V\}$$ This is also naturally identified with the associated vector bundle: $$V_n^k\times_\rho \mathbb R^k$$ where $\rho$ is the standard representation of $O(k)$ on $\mathbb R^k$.

We define the universal bundle as the direct limit: $$\gamma_k=\varinjlim_{n\geq k}\gamma_k^n$$ and I am trying to show that it is a vector bundle over the classifying $BO_k=\varinjlim_{n\geq k}Gr_k(\mathbb R^n)$. Using the universal property of the direct limit it was easy to show that there is a well defined continuous surjective map $\pi:\gamma_k\rightarrow BO_k$. I can also show that $\pi^{-1}([V])$ is isomorphic to $\mathbb R^k$ by appealing to the universal property of the direct limit.

I have tried to do something similar to show that the $\gamma_k$ is locally trivially, but I am struggling. In particular, let $\phi_n:Gr_k(\mathbb R^n)\rightarrow BO_k$ be the maps defining the topology on $BO_k$, $\psi_n$ be the same maps for $\gamma_k$, and $\pi_n$ ve the projection for each $\gamma_k^n\rightarrow Gr_k(\mathbb R^n)$. Then I take neighborhood $U$ of $[V]$ in $BO_k$, so $\pi^{-1}(U)$ is open in $\gamma_k$ implying that for all $n$ we have that $\psi_n^{-1}(\pi^{-1}(U))=\pi_n^{-1}(\phi_n^{-1}(U))$ is open in $\gamma_k^n$. Shrinking $U$ if necessary I believe we can take each of these to be trivialization of $\gamma_n^k$. But then I need my charts to commute inclusions, and I am not sure how to do that, so I am a bit stuck.

Any help would be appreciated. I know this is treated in Milnor's characteristic classes, but since so many modern sources define this object explicitly as a direct limit I reckon there should be a way to show it using the definition of the topology on the space.

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  • $\begingroup$ I don’t think it’s difficult to show that the definition in Milnor—Stasheff is equivalent to the direct limit definition. $\endgroup$
    – Frank
    Commented May 12 at 5:44
  • $\begingroup$ @Frank thank you I have figured it out. $\endgroup$
    – Chris
    Commented May 12 at 11:21

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